Notes of Bernard Maurey’s lecture nr 2

1. More examples of type and cotype

Let {X} be a Banach space. The possible types are in {I_T =[1,p_X)} or {I_T =[1,p_X]}, {pX \leq 2}. The possible cotypes are in {I_C =[q_X ,\infty)} or {I_C =(q_X ,\infty)}, {q_X \geq 2}.

Cotype does not pass to quotients. Indeed, {\ell_1} has cotype {2}, and every separable Banach space is a quotient of {\ell_1}. For instance, {c_0} has trivial cotype.

The type of {L_p (X)} is {\min\{type(L_p),type(X)\}}.

1.1. Complex interpolation

Type {p} is equivalent to the map {\ell_p (X)\rightarrow L_p (X)}, {(x_i)\mapsto \sum\epsilon_i x_i} being bounded. By interpolation, if {X_0} has type {p_0} and {X_1} has type {p_1}, the interpolation space {X_{\theta}}, {\theta\in (0,1)}, has type {p} such that

\displaystyle  \begin{array}{rcl}  \frac{1}{p}=\frac{1-\theta}{p_0}+\frac{\theta}{p_1}. \end{array}

Cotype does not work so nicely. It is expressible in terms of a map {Rad(X)\rightarrow\ell_q (X)}, where

\displaystyle  \begin{array}{rcl}  Rad(X)=\{\sum\epsilon_i x_i \,;\,x_i \in X,\,\epsilon_i \in\{\pm 1\}\}\}\subset L_q (X). \end{array}

If {Rad(X)} is complemented, i.e. is the image of a projector {L_q (X)\rightarrow Rad(X)} independent on {q} (this is the notion of {K}-convexity), interpolation works for cotype.

1.2. Super

Clearly, type and cotype are super properties. A stronger property holds.

Definition 1 Say {X} is crudely finitely representable in {Y} if there exists {C} such that every finite dimensional subspace of {X} embeds with distorsion {\leq C} in {Y}.

The same proof as for ordinary finite representability gives

Proposition 2 {X} is crudely finitely representable in {Y} iff {X} is isomorphic to a subspace of an ultrapower {Y^I /\mathcal{u}}.

1.3. Type and finite representability

Theorem 3 (Pisier, Maurey-Pisier 1973) Let {X} be a Banach space. Then {X} has trivial type iff {\ell_1} is finitely representable in {X}. {X} has trivial cotype iff {c_0} is finitely representable in {X}.

Lemma 4 Let {Y_1 ,\ldots,Y_n} be independent {X}-valued random variables with a symmetric distribution. Let {\epsilon_i} be independent Bernoulli variables. Then {(\epsilon_1 Y_1 ,\ldots,\epsilon_n Y_n)} and {Y_1 ,\ldots,Y_n} have the same distribution.

Proof: of Theorem 3. Let {a_n} denote the best constant in the inequality

\displaystyle  \begin{array}{rcl}  (\mathop{\mathbb E}|\sum_{i=1}^{n}\epsilon_i x_i|^2 )^{1/2} \leq \sqrt{n}a_n (\sum|x_i|^2 )^{1/2} . \end{array}

Then {a_n \leq 1}. Let us show that {(a_n)} is submultiplicative, i.e. {a_{mn}\leq a_m a_n}. Indeed, group {mn} variables in {m} packages of {n}.

Assume that {a_n =1} for all {n}. So for every {n}, there exist points {\mathop{\mathbb E}|\sum_{i=1}^{n}\epsilon_i x_i|^2} is close to {a_n^2 (\sum|x_i|^2 )}. Go to an ultrapower to find vectors {\xi_i} such that {\mathop{\mathbb E}|\sum_{i=1}^{n}\epsilon_i \xi_i|^2 =a_n^2 (\sum|\xi_i|^2 )}. The equality case in the chain of inequalities leading to {a_n \leq 1} shows that all norms {|\xi_i |=1} and for all choices of signs, {|\sum\pm \xi_i|=n}. Fix signs {\epsilon_i}. If {x^*} is a unit functional, {x^*(\sum\epsilon_i \xi_i)=n}, thus {x^* (\xi_i)=\epsilon_i}. In other words, these functionals give an isometry of span{(\x_i)} onto {\ell_1}.

Otherwise, there exists {n_0} such that {a_{n_0}<1}. Then there exists {C>0}, {\alpha>0}, such that {a_n \leq C n^{-\alpha}}. Let {p\geq p_X}. Given a family {x_i \in X}, split it according to norms,

\displaystyle  \begin{array}{rcl}  A_k =\{i\,;\,2^{-k-1}<|x_i|\leq 2^{-k}\}. \end{array}

Then

\displaystyle  \begin{array}{rcl}  (\mathop{\mathbb E}|\sum_{i\in A_k}\epsilon_i x_i|^2 )^{1/2} \leq \sqrt{|A_k|}a_{|A_k|}2^{-k}\sqrt{|A_k|}. \end{array}

Summing over {k} gives the wanted estimate. \Box

1.4. Ribe’s Theorem

It is our first nonlinear result.

Theorem 5 (Ribe 1976) If {T:X\rightarrow Y} is a uniform homeomorphism, then {X} is crudely finitely representable in {Y} and {Y} is crudely finitely representable in {X}.

Proof: Upgrade the uniform homeomorphism to a bi-Lipschitz map. Use differentiability on finite dimensional subspaces to improve bi-Lipschitz maps to isomorphisms. \Box

This implies that {L_p [0,1]} and {L_q [0,1]} are not uniformly homeomorphic, an earlier result due to Enflo. Enflo’s proof relies on a metric notion of type, on which Naor and Schechtman, Mendel and Naor have elaborated since.

Definition 6 Let {E} be a metric space. Say {E} has Enflo type {p} if there is a constant {C>0} such that for all {f:\{\pm 1\}^n\rightarrow E} ,

\displaystyle  \begin{array}{rcl}  \mathop{\mathbb E}_{\epsilon}d(f(\epsilon),f(-\epsilon))^p \leq C^p \sum_{j=1}^{n}\mathop{\mathbb E}_{\epsilon}d(f(\epsilon),f(R_j \epsilon))^p , \end{array}

where {R_j} changes the sign of the {j}-th coordinate.

Remark 1 For a Banach space, Enflo type {p} implies usual type {p}.

1.5. Proof of Ribe’s Theorem

Let {T:X\rightarrow Y} be a uniform homeomorphism. Then {T} is large-scale Lipschitz: {|T x_1 -T x_2|\leq C|x_1 -x_2|} provided {|x_1 -x_2|>1}. Go to an ultra product to get rid the the large-scale restriction. This means setting {\tilde{T}x =(T(kx)/k)_{k\in{\mathbb N}}\in Y^{{\mathbb N}}}. This is bi-Lischitz: {X\rightarrow Y^{{\mathbb N}}_{\mathcal{U}}}. On every finite dimensional subspace {X_0 \subset X}, {\tilde{T}} has points of differentiability. The differential is a linear isomorphism of {X_0} onto a linear subspace of {Y^{{\mathbb N}}_{\mathcal{U}}}. So {X} is coarsely finitely representable in {Y^{{\mathbb N}}_{\mathcal{U}}}, and thus in {Y}.

2. {B}-convexity

This goes back to Beck and Giesy. They were lookibg for a law of large numbers for Banach space valued random variables.

Lemma 7 Let {B} have type {p>1}. Let {X_1 ,\ldots,X_n ,\ldots} be independent centered {B}-valued random variables. Then

\displaystyle  \begin{array}{rcl}  \mathop{\mathbb E}|\sum_{i=1}^{n}X_i|^p \leq 2^{p}T_{p}^{p}\sum_{i=1}^{n}\mathop{\mathbb E}|X_i |^p. \end{array}

Proof: Reduce to the case of symmetric distributions by replacing {X_i} by {X_i -X'_i} where {X'_i} is an independent copy of {X_i}. This produces the factor {2^p}. Then apply Jensen’s inequality

\displaystyle  \begin{array}{rcl}  \mathop{\mathbb E}|\sum_{i=1}^{n}X_i|^p \leq\mathop{\mathbb E}|\sum_{i=1}^{n}X_i|^p \end{array}

\Box

It implies that

\displaystyle  \begin{array}{rcl}  \mathop{\mathbb E}|\frac{1}{n}\sum_{i=1}^{n}X_i|\leq(\mathop{\mathbb E}|\frac{1}{n}\sum_{i=1}^{n}X_i|^p)^{1/p}\leq 2T_p n^{(1-p)/p} ,\end{array}

which tends to {0}, this is a law of large numbers.

2.1. Application: approximation of convex hulls

Corollary 8 Let {B} be a Banach space with type {p>1}. Let {A\subset U_B}. Let {x\in Conv(A)}. There exist {a_1 ,\ldots,a_n \in A} such that

\displaystyle  \begin{array}{rcl}  |s-\frac{1}{n}\sum_{i=1}^{n}a_j|\leq 4 T_p n^{(1-p)/p}. \end{array}

Proof: Apply the law of large numbers to independent copies of the following random variable {U}. By assumption, there exist {\lambda_i \geq 0}, {\sum\lambda_i =1} such that {x=\sum\lambda_i x_i}. So let {U=x_i} with probability {\lambda_i}. \Box

3. Factorization

The following was one of the first results on isomorphism types.

Theorem 9 (Kwapién) If a Banach space has simultaneously type {2} and cotype {2}, then is is isomorphic to a Hilbert space.

Up to now we have uses Bernoulli averages. Why not use Gaussian averages ? Grouping {mn} Bernoulli variables (divided by {n}) in packages of {n} produces {m} variables which are approximately standard Gaussian (central limit theorem). So we can use Gaussian averages instead.

Lemma 10 Let {G_1}, {G_2} be {E}-valued centered Gaussian vectors. Assume that for all {t\in E^*},

\displaystyle  \begin{array}{rcl}  \mathop{\mathbb E}(tG_1)^2 :=Q_1 (t)\leq Q_2 (t):=\mathop{\mathbb E}(tG_2)^2 . \end{array}

Then for every convex function {f},

\displaystyle  \begin{array}{rcl}  \mathop{\mathbb E}(f(G_1))\leq \mathop{\mathbb E}(f(G_2)). \end{array}

Proof: Pick a common orthogonal basis for the quadratic forms {Q_1} and {Q_2}. Use such a basis to find {G'} independant from {G_1} such that both {G_1 \pm G'} have the same distribution as {G_1}. \Box

Let {E} be a finite dimensional normed space. Say a homogeneous function {\phi:E\rightarrow{\mathbb R}_+} is a cotype {2} function if for all {x_1 ,\ldots,x_n \in E},

\displaystyle  \begin{array}{rcl}  \sum\phi(x_i)^2 \leq C_{\phi}^2 \mathop{\mathbb E}|\sum g_i x_i|^2 \end{array}

where {g_i} are independent standard Gaussians.

Lemma 11 There exists a measure on the unit sphere {S} of {S^*} such that for all {x\in E},

\displaystyle  \begin{array}{rcl}  \frac{1}{C_{\phi}}\phi(x)\leq(\int_{S}(t\cdot x)^2 \,d\mu(t))^{1/2}\leq T_{2,E}|x|. \end{array}

Proof: Set {f_x (t)=\sum(t\cdot x_i)^2},

\displaystyle  \begin{array}{rcl}  C_1=\{f_x -g \,;\, \sum|x_i|^2<T_{2}^{-2},\,g>0\textrm{ on }S\}, \end{array}

\displaystyle  \begin{array}{rcl}  C_2=\{f_y \,;\,\sum|y_i|^2 \geq C_{\phi}^{2}\}. \end{array}

These are disjoint convex sets \Box

I have to run to catch the 4:30 train to Orsay…

About metric2011

metric2011 is a program of Centre Emile Borel, an activity of Institut Henri Poincaré, 11 rue Pierre et Marie Curie, 75005 Paris, France. See http://www.math.ens.fr/metric2011/
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